Two small pith balls, each of mass m = 11.2 g, are suspended from the ceiling of the physics lab by 1.7 m long fine strings and are not moving. If the angle which each string makes with the vertical is è = 30.4°, and the charges on the two balls are equal, what is the magnitude of that charge (in µC)? (Please note that the unit of charge is micro-Coulomb here - some browsers might not display the Greek letter mu correctly and show it as an m.)
If the charged ball is suspended be the string which is deflected by the angle α, the forces acting on it are: mg (downwards), tension T (along the string - to the pivot point), and F (electric force –along the line connecting the charges).
Projections on the horizontal and vertical axes are:
x: T•sin α = F, ….(1)
y: T•cosα = mg. ….(2)
Divide (1) by (2):
T•sin α/ T•cosα = F/mg,
tan α = F/mg.
Since
q1=q2=q.
r=2•L•sinα,
k=9•10^9 N•m²/C²
F =k•q1•q2/r² = k•q²/(2•L•sinα)².
tan α = F/mg =
= k•q²/(2•L•sinα)² •mg.
q = (2•L•sinα) • sqrt(m•g•tanα/k)=
=(2•1.7•0.506) •sqrt(0.0112•9.8•0.577/9•10^9) =
=4.5•10^-6 C =4.5 μC.
Well, it looks like we're in for an electrifying problem! Let's zap straight into it!
We have two pith balls with a mass of 11.2 g each. They're suspended from the ceiling, creating a hair-raising scene! The length of the strings is 1.7 m, but the real shocker is the angle each string makes with the vertical: 30.4 degrees!
To find the magnitude of the charge on the pith balls, we need to do a bit of electrostatics hocus-pocus. Here's the formula we'll use:
q = (mg) / (k * tan(è))
Where:
- q is the magnitude of the charge (what we're looking for)
- m is the mass of the pith balls (11.2 g)
- g is the acceleration due to gravity (a shocking 9.8 m/s^2)
- k is Coulomb's constant (k = 9 x 10^9 Nm^2/C^2)
- è is the angle the strings make with the vertical (30.4 degrees)
Plugging in the givens, we get:
q = (0.0112 kg * 9.8 m/s^2) / (9 x 10^9 Nm^2/C^2 * tan(30.4 degrees))
After doing the math, the magnitude of the charge on each pith ball is approximately 5.57 µC (micro-Coulombs)!
So, the pith balls are charged up and ready to go! Time to bring on the science circus!
To find the magnitude of the charge on the pith balls, we can use Coulomb's Law and the principle of electrostatic equilibrium.
1. Start by finding the tension force in one of the strings. The tension force in the string is equal to the weight of the pith ball, since the ball is in equilibrium.
Tension = Weight of the pith ball = mass * gravitational acceleration
Tension = 11.2 g * 9.8 m/s²
Tension = 0.10976 N
2. Since the angle between the string and the vertical is given as è = 30.4°, we can decompose the tension force into two components: one along the vertical direction and the other along the horizontal direction.
Vertical component = Tension * cos(è)
Vertical component = 0.10976 N * cos(30.4°)
Horizontal component = Tension * sin(è)
Horizontal component = 0.10976 N * sin(30.4°)
3. The electrostatic force between the two pith balls is equal to the horizontal component of tension in either string. Since the charges on the balls are equal, the magnitude of the electrostatic force between them is:
Electrostatic force = Horizontal component = 0.10976 N * sin(30.4°)
4. According to Coulomb's Law, the electrostatic force between two charges is given by the equation:
Electrostatic force = (k * |q₁ * q₂|) / r²
Where:
- k is Coulomb's constant, approximately 9 × 10^9 N m²/C²
- q₁ and q₂ are the magnitudes of the charges
- r is the distance between the charges
In this case, the distance between the charges is equal to the length of the string, 1.7 m. Rearranging the equation, we can solve for the magnitude of the charge:
|q₁ * q₂| = (Electrostatic force * r²) / k
|q₁ * q₂| = (0.10976 N * sin(30.4°) * (1.7 m)²) / (9 × 10^9 N m²/C²)
|q₁ * q₂| = 4.49665 x 10^-10 C²
5. Since the charges on the pith balls are equal, we can take the square root of the above value to find the magnitude of each charge:
|q| = √(4.49665 x 10^-10 C²)
|q| = 6.70 x 10^-6 C
Therefore, the magnitude of the charge on each pith ball is 6.70 µC.
To find the magnitude of the charge on the pith balls, we can use the concept of electrostatic force and equilibrium.
First, let's discuss the forces acting on each pith ball. There are two forces acting on each ball:
1. The force due to gravity, which is equal to the weight of the ball (mg), where m is the mass of the ball and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The electrostatic force between the two charged pith balls. Since the charges on the balls are equal, the electrostatic force will be repulsive.
We can use the fact that the system is in equilibrium, which means the net force on each ball is zero. This implies that the weight of the ball is balanced by the electrostatic force.
The electrostatic force between the two balls can be expressed as:
F = k * (q^2 / r^2)
where F is the force, k is the electrostatic constant (approximately 9 × 10^9 N·m^2/C^2), q is the magnitude of the charge on each ball, and r is the distance between the centers of the balls.
Since the balls are not moving and are in equilibrium, the net force on each ball is zero. This means that the weight of each ball is equal to the electrostatic force. We can equate these forces:
mg = k * (q^2 / r^2)
Now, let's solve for the charge q.
1. Convert the mass of the pith ball to kilograms:
m = 11.2 g = 0.0112 kg
2. Convert the angle to radians:
θ = 30.4° * π/180 = 0.531 rad
3. Find the distance between the centers of the balls using the given length and angle:
r = 1.7 m * sin(θ) = 1.7 m * sin(0.531) ≈ 1.19 m
4. Substitute the known values into the equation and solve for q:
q = √((mg * r^2) / k)
q = √((0.0112 kg * 9.8 m/s^2 * (1.19 m)^2) / (9 * 10^9 N·m^2/C^2))
q ≈ 0.000607 C ≈ 607 µC
Therefore, the magnitude of the charge on each pith ball is approximately 607 micro-Coulombs (µC).